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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Migratory Birds
  5. Discussions

Migratory Birds

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3628 Discussions

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  • TopTalent
     + 0 comments

    Java

    public static int migratoryBirds(List<Integer> arr) {
        int max = Integer.MIN_VALUE;
        int count = 0;
        int idMin = Integer.MAX_VALUE;
        int id = 0;
        
        Collections.sort(arr);
        
        for (int i = 0, j = 1; j < arr.size() - 1; i++, j++) {
            if (arr.get(i) == arr.get(j + 1)) {
                id = arr.get(i);
                count++;
                if (count > max) {
                    max = count;
                    idMin = id;
                }
            } else {
                count = 0;
            }
        }
        return idMin;
    }

  • arthur_frachon
     + 0 comments

    Fast and easy : Python3

    def migratoryBirds(arr):
    # Write your code here
    dic = {key:arr.count(key) for key in set(arr)}
    maxKey = [key for key,val in dic.items() if val == max(dic.values())]
    return min(maxKey)

  • tongyou1995
     + 0 comments

    My Python 3 O(n) solution, iterating once through array to build dictionary or hash map, then count the keys that have a value equal to maximum and append those keys to an array. If more than one key exists, the minimum does nothing:

    import os
def migratoryBirds(arr):
    freq = {}
    for val in arr:
        freq[val] = freq.get(val, 0) + 1
    max_val = max(freq.values())
    keys = []
    for k, v in freq.items():
        if v == max_val:
            keys.append(k)
    return min(keys)
        
if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    arr_count = int(input().strip())
    arr = list(map(int, input().rstrip().split()))
    result = migratoryBirds(arr)
    
    fptr.write(str(result) + '\n')
    fptr.close()

  • rezl_jaws
     + 0 comments

    Here is my simple C++ solution,

    int migratoryBirds(vector arr) {

    int cnt = 0;
int prevCnt = 0;
int res  = 0;
for( int i =1; i <=5; i++)  // if you read the constraint it says only 1 - 5 types guranteed, so we find the max of each element and the corresponding min
{
    cnt = std::count (arr.begin(), arr.end(), i);

    if( prevCnt < cnt )
    {
        prevCnt = cnt;
        res = i;
    }
    else if( prevCnt == cnt )
    {
        res = min(res, i); // here we get the min of the max cnt if they are the same
    }
}
return res;

    }

  • Fas5ter
     + 0 comments

    Here is my Python solution

    def migratoryBirds(arr):
    # Write your code here
    
    min_number = min(arr)
    max_number = max(arr)

    # Find the lowest type id of the most frequently sighted birds
    lowestId = max_number
    for i in range(max_number-1, min_number, -1):
        x = arr.count(i)
        if x >= (arr.count(lowestId)):
            lowestId = i
    return lowestId